(i) Find the gradient of line AB:
The gradient \(m\) of line \(AB\) is given by:
\(m = \frac{3a + 9 - (2a - 1)}{2a + 4 - a} = \frac{a + 10}{a + 4}\)
The gradient of a line perpendicular to \(AB\) is the negative reciprocal:
\(m_{\perp} = -\frac{a + 4}{a + 10}\)
(ii) Find the possible values of \(a\):
The distance \(AB\) is given by:
\(\sqrt{(2a + 4 - a)^2 + (3a + 9 - (2a - 1))^2} = \sqrt{260}\)
Simplify the expression:
\(\sqrt{(a + 4)^2 + (a + 10)^2} = \sqrt{260}\)
Square both sides:
\((a + 4)^2 + (a + 10)^2 = 260\)
Expand and simplify:
\(a^2 + 8a + 16 + a^2 + 20a + 100 = 260\) \(2a^2 + 28a + 116 = 260\) \(2a^2 + 28a - 144 = 0\)
Divide by 2:
\(a^2 + 14a - 72 = 0\)
Factorize:
\((a - 4)(a + 18) = 0\)
Thus, \(a = 4\) or \(a = -18\).